你加上 WHERE 要筛选的子句。

SELECT t.diapason,
       COUNT(*) AS 'number_of_users'
FROM (
    SELECT 
        CASE
            WHEN amount < 200 THEN '0-200'
            WHEN amount >= 200 THEN '200 +'
        END AS diapason
    FROM   (
        SELECT payments.user_id, SUM(amount) AS amount
        FROM   payments
        INNER JOIN activity ON payments.user_id = activity.user_idAND activity.login_time = payments.payment_time
        WHERE activity.login_time BETWEEN '2018-01-10' AND '2018-01-12'
        GROUP  BY payments.user_id
    ) p
) t
GROUP  BY t.diapason
ORDER  BY number_of_users DESC;

我理解你的问题是这样的。在2018年1月1日至2018年1月12日期间，您有3个用户（user_id=1,2,3）登录。在这些用户中，用户id 1支付了2笔共250，用户id 2支付了1笔共40，用户id 3支付了0，因此他们的总金额为0。因此，范围内有2个值 0-200 ，1在范围内 200 + . 如果这是正确的理解，此查询将为您提供所需的结果：

SELECT CASE  
         WHEN amount < 200 THEN '0-200'
         WHEN amount >= 200 THEN '200 +'
       END AS diapason,
       COUNT(*) AS number_of_users
FROM (SELECT a.user_id, COALESCE(SUM(p.amount), 0) AS amount
      FROM activity a
      LEFT JOIN payments p ON p.user_id = a.user_id
      WHERE a.login_time BETWEEN '01.01.2018' AND '12.01.2018'
      GROUP BY a.user_id) p
GROUP BY diapason;

输出：

diapason    number_of_users
0-200       2
200 +       1

sqlfiddle演示
更新
在总计中添加另一行 number_of_users ，只需添加 WITH ROLLUP 到 GROUP BY 条款：

SELECT CASE  
         WHEN amount < 200 THEN '0-200'
         WHEN amount >= 200 THEN '200 +'
       END AS diapason,
       COUNT(*) AS number_of_users
FROM (SELECT a.user_id, COALESCE(SUM(p.amount), 0) AS amount
      FROM activity a
      LEFT JOIN payments p ON p.user_id = a.user_id
      WHERE a.login_time BETWEEN '01.01.2018' AND '12.01.2018'
      GROUP BY a.user_id) p
GROUP BY diapason WITH ROLLUP

输出：

diapason    number_of_users
0-200       2
200 +       1
(null)      3

在应用程序框架中，可以使用 diapason 价值是 NULL 输出类似 Total 相反。
更新的sqlfiddle
在mysql中也可以做同样的事情（参见这个sqlfiddle），方法是将这个查询 Package 成子查询并使用 COALESCE 上 diapason 列。在这种情况下，输出为：

diapason    number_of_users
0-200       2
200 +       1
Total       3